Answer
(a) $x=0$ is a vertical asymptote.
$y = 1$ is a horizontal asymptote.
(b) $f$ is decreasing on the intervals $(-\infty,0)\cup (2, \infty)$
$f$ is increasing on the interval $(0,2)$
(c) The local maximum is $f(2) = \frac{5}{4}$
(d) The graph is concave down on these intervals: $(-\infty,0)\cup (0,3)$
The graph is concave up on this interval: $(3, \infty)$
The point of inflection is $(3,\frac{11}{9})$
(e) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = 1+\frac{1}{x}-\frac{1}{x^2}$
Note that $x \neq 0$.
$x=0$ is a vertical asymptote.
$\lim\limits_{x \to -\infty}f(x) = 1$
$\lim\limits_{x \to \infty}f(x) = 1$
$y = 1$ is a horizontal asymptote.
(b) We can find the points where $f'(x) = 0$:
$f'(x) = -\frac{1}{x^2}+\frac{2}{x^3} = 0$
$\frac{1}{x^2} = \frac{2}{x^3}$
$x^3 = 2x^2$
$x = 2$
Note that $f'(x)$ is not defined at the point $x=0$
When $x \lt 0~~$ or $x \gt 2~~$ then $f'(x) \lt 0$
$f$ is decreasing on the intervals $(-\infty,0)\cup (2, \infty)$
When $x \gt 0$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(0,2)$
(c) $f(2) = 1+\frac{1}{2}-\frac{1}{2^2} = \frac{5}{4}$
The local maximum is $f(2) = \frac{5}{4}$
There is no local minimum.
(d) We can find the points where $f''(x) = 0$:
$f''(x) = \frac{2}{x^3}-\frac{6}{x^4} = 0$
$\frac{2}{x^3} = \frac{6}{x^4}$
$2x^4 = 6x^3$
$x = 3$
Note that $f''(x)$ is undefined at $x = 0$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on these intervals: $(-\infty,0)\cup (0,3)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(3, \infty)$
$f(3) = 1+\frac{1}{3}-\frac{1}{3^2} = \frac{11}{9}$
The point of inflection is $(3,\frac{11}{9})$
(e) We can see a sketch of the graph below.
